How do you simplify #sqrt(6) *sqrt(32)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Apr 29, 2015 Remember that #sqrt(a)*sqrt(b)=sqrt(a*b)# So: #sqrt(6)*sqrt(32)=sqrt(6*32)=sqrt(192)=sqrt(4*4*4*3)=2*2*2*sqrt(3)=8sqrt(3)# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 2773 views around the world You can reuse this answer Creative Commons License